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2 years ago

Fin each missing measure. Round all answers to the nearest tenth. Pls Help!!

Mathematics

1 answer:

stiv31 [10]2 years ago

8 0

Answer:

6.2 units

Step-by-step explanation:

The mnemonic SOH CAH TOA is intended to remind you of the relations between trig functions and sides of a right triangle.

<h3>Common segment</h3>

To find the value of x in this figure, we need to know the length of the common segment between the two triangles. That segment is opposite the 53° angle, so can be found using the sine relation:

Sin = Opposite/Hypotenuse

Opposite = Hypotenuse × Sin = 30·sin(53°) ≈ 23.9591

<h3>Missing measure</h3>

The side marked x is adjacent to the marked angle in that triangle, so the relevant relation is ...

Cos = Adjacent/Hypotenuse

Adjacent = Hypotenuse × Cos = 23.9591·cos(75°) ≈ 6.20106

The length of the side marked x is about 6.2 units.

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4 years ago

Can you help me please to prove that the bleu areas are same without without useing numbers.

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2 years ago

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$

3 0

2 years ago

Please help!! I’ll mark brainliest!

elena-14-01-66 [18.8K]

Answer:

Just from looking at the picture, I'm assuming that we're trying to find the terms that are like to the terms in the blue section.

1. 8c since both are in terms of c

2. 7t since both are in terms of t

3. 8r since both are in terms of r

4. -4n since both are in terms of n

5. xy since both are in terms of xy

6. 16 since both don't have any variables

7. 4sp since both are in terms of sp

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Write the equation of the line that passes through (−3, 5) and (2, 10) in slope-intercept form.

Paladinen [302]

Answer:

gradient =y2-y1÷x2-x1

=10-5÷2-(-3)

y-y1=m(x-x1)

y-5=1(x-(-3))

y=x+2

7 0

3 years ago